You have found the following ages (in years) of all 6 lions at your local zoo: $ 8,\enspace 1,\enspace 2,\enspace 10,\enspace 8,\enspace 17$ What is the average age of the lions at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 6 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{8 + 1 + 2 + 10 + 8 + 17}{{6}} = {7.7\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $8$ years $0.3$ years $0.09$ years $^2$ $1$ year $-6.7$ years $44.89$ years $^2$ $2$ years $-5.7$ years $32.49$ years $^2$ $10$ years $2.3$ years $5.29$ years $^2$ $8$ years $0.3$ years $0.09$ years $^2$ $17$ years $9.3$ years $86.49$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.09} + {44.89} + {32.49} + {5.29} + {0.09} + {86.49}} {{6}} $ $ {\sigma^2} = \dfrac{{169.34}}{{6}} = {28.22\text{ years}^2} $ The average lion at the zoo is 7.7 years old. The population variance is 28.22 years $^2$.